Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
template<class T> inline T sqr(T x) {return x*x;}

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long long ll;

#define all(a)  (a).begin(),(a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define pb push_back
#define mp make_pair
#define each(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define exist(s,e) ((s).find(e)!=(s).end())
#define range(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)  range(i,0,n)
#define clr(a,b) memset((a), (b) ,sizeof(a))
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

const double eps = 1e-10;
const double pi  = acos(-1.0);
const ll INF =1LL << 62;
const int inf =1 << 29;

const int vmax=100010;
int sdist[vmax];
int tdist[vmax];

ll snum[vmax];
ll tnum[vmax];

vi graph[vmax];

int main(void){
	int n,m,s,t;
	cin  >> n >> m >> s >> t;
	s--,t--;

	rep(i,m){
		int x,y;
		cin >> x >> y;
		x--,y--;
		graph[x].pb(y);
		graph[y].pb(x);
	}
	rep(i,vmax) sdist[i]=tdist[i]=-1;

	queue<pii> q;
	sdist[s]=0; q.push(mp(s,0));
	while(!q.empty()){
		pii cur=q.front();q.pop();
		int v=cur.first;
		int c=cur.second;
		for(auto &v2:graph[v]){
			if(sdist[v2]==-1){
				sdist[v2]=c+1;
				q.push(mp(v2,c+1));
			}
		}
	}

	tdist[t]=0; q.push(mp(t,0));
	while(!q.empty()){
		pii cur=q.front();q.pop();
		int v=cur.first;
		int c=cur.second;
		for(auto &v2:graph[v]){
			if(tdist[v2]==-1){
				tdist[v2]=c+1;
				q.push(mp(v2,c+1));
			}
		}
	}
	rep(i,n) snum[sdist[i]]++;
	rep(i,n) tnum[tdist[i]]++;
	int sp=sdist[t];
	ll ans=0LL;
	rep(i,sp-1) ans+=snum[i]*tnum[sp-2-i];
	cout << ans << endl;
	return 0;
}